A marshmallow is placed in a microwave to be heated for smores. The marshmallow has a 86 mL at a temperature of 543 °C. What temperature, in K, would themarshmallow need to be if the volume changed to 51.1 ml?
Answers
Answer: 484.94 K
Explanation:
T2 = T1 X V2 / V1
Temperature must be in kelvin so 543 + 273.15 =816.15
816.15 X 51.1 / 86
Answer: We can use the combined gas law to solve this problem:
(P₁V₁/T₁) = (P₂V₂/T₂)
where P is pressure, V is volume, and T is temperature in Kelvin.
We know that P₁ = P₂ (the pressure is assumed to be constant), and we are given V₁, T₁, and V₂. We can solve for T₂:
(P₁V₁/T₁) = (P₂V₂/T₂)
T₂ = (P₂V₂/T₁) * (T₁/P₁V₁)
We need to convert the initial temperature from Celsius to Kelvin:
T₁ = 543 + 273 = 816 K
Substituting the values:
T₂ = (1 atm * 86 mL / 816 K) * (51.1 mL / 1 atm * 86 mL)
T₂ = 0.0629 * 51.1 * 1000 = 3217 K
Therefore, the marshmallow would need to be heated to a temperature of 3217 K for its volume to change from 86 mL to 51.1 mL.
Enjoy (:
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Answer:
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I hope this helps. Let me know in the comments if anything is unclear.